适用于List`dist`功能的分销商 [英] Applicative Distributor for List `dist` Function

问题描述

我认为我从 dist 函数。 htmlrel =nofollow>使用效果编程：

本文在前面加上这个函数：

你是否注意到序列和转置现在看起来相似？

编译器从它们的类型中推断出区分这两个程序的细节。两者都是应用型
分销商的实例

  dist :: Applicative f => [f a]  - > f [a] 
 dist [] = [[]] 
 dist（v：vs）= [（:) v（dist vs）] 
  pre> 
 
 
但是，我得到以下编译时错误： 
 
 
  ghci中> ：l ApplicativePaper.hs 
 [1的1]编译Main（ApplicativePaper.hs，解释）
 
 ApplicativePaper.hs：12：20：
无法推导出（a〜fa ）
来自上下文（Applicative f）
由类型签名绑定
 dist :: Applicative f => [f a]  - > f [a] 
在ApplicativePaper.hs：10：9-39 
`a'是一个刚性类型变量，由
绑定dist :: Applicative f =>的类型签名。 [f a]  - > f [a] 
在ApplicativePaper.hs：10：9 
相关绑定包括
 vs :: [fa]（绑定在ApplicativePaper.hs：12：9）
v :: fa（bound at ApplicativePaper.hs：12：7）
 dist :: [fa]  - >在[（:)的第一个参数中，即'v'
在表达式中：（:) v（dist vs）（在ApplicativePaper.hs：11： 
失败，模块加载：无。 
  
请让我知道我做错了什么。另外，请为此功能提供直觉。
解决方案
在本文中，双括号 [f u1 ... un = 在第4页定义为与纯f <*>相同。 u1 * ...< *> un 。
 
 
 
 $  dist 在文本中的定义是
  dist ::应用f⇒[fa]→f [a] 
 dist [] = [[] = 
 dist （v：vs）= [（:) v（dist vs）= 
  
将其他符号翻译成Haskell并取代 [... = 的定义导致
  dist :: Applicative f => [f a]  - > f [a] 
 dist [] =纯[] 
 dist（v：vs）=纯（:) * v * （dist vs）
  
 
I think that I copied the following dist function from Applicative Programming with Effects:

The paper prefaces this function with:

  
Have you noticed that sequence and transpose now look rather alike?
  The details that distinguish the two programs are inferred by the
  compiler from their types. Both are instances of the applicative
  distributor for lists


dist :: Applicative f => [f a] -> f [a]
dist []       = [[]]
dist (v : vs) = [(:) v (dist vs)]
However, I get the following compile-time error:
ghci> :l ApplicativePaper.hs
[1 of 1] Compiling Main             ( ApplicativePaper.hs, interpreted )

ApplicativePaper.hs:12:20:
    Could not deduce (a ~ f a)
    from the context (Applicative f)
      bound by the type signature for
                 dist :: Applicative f => [f a] -> f [a]
      at ApplicativePaper.hs:10:9-39
      `a' is a rigid type variable bound by
          the type signature for dist :: Applicative f => [f a] -> f [a]
          at ApplicativePaper.hs:10:9
    Relevant bindings include
      vs :: [f a] (bound at ApplicativePaper.hs:12:9)
      v :: f a (bound at ApplicativePaper.hs:12:7)
      dist :: [f a] -> f [a] (bound at ApplicativePaper.hs:11:1)
    In the first argument of `(:)', namely `v'
    In the expression: (:) v (dist vs)
Failed, modules loaded: none.
Please let me know what I'm doing wrong. Also, please provide an intuition for this function.
解决方案
In the paper, the double bracket 〚 f u1 … un 〛 is defined on page 4 to be the same as pure f <*> u1 <*> … <*> un.

The definition of dist in the text is
dist :: Applicative f ⇒ [f a] → f [a ]
dist []       = 〚 [] 〛
dist (v : vs) = 〚 (:) v (dist vs) 〛
Translating the other symbols into Haskell and making the substitution for the definition of 〚 … 〛 results in
dist :: Applicative f => [f a] -> f [a ]
dist []       = pure []
dist (v : vs) = pure (:) <*> v <*> (dist vs)


                        
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